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Alex Ar [27]
3 years ago
10

Find the area of triangle CDE the length of the square is 24 meters;

Mathematics
1 answer:
Aleksandr [31]3 years ago
6 0

Answer:

288

Step-by-step explanation:

If you didnt know a triangle has the area of a square/2

so basically what im saying is the area of that square/2 equals the area of the triangle

to find the area of the square you take 24 and square it

24^2=576

finally we divide that by 2

576/2=288

so the area of the triangle is 288

have a great day :D

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Solve using inverse operations 83+q=125; q=
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Answer:

42

Step-by-step explanation:

83+q=125

q=125-83

q=42

3 0
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Someone please help!! I’ll give brainliest!
snow_lady [41]

Answer:

1766.3

Step-by-step explanation:

V=4 /3πr^3

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3 years ago
rotation of 90 degrees counterclockwise about the origin, point O, then a reflection across the x-axis reflection across the y-a
SpyIntel [72]

Answer:

The point O can be (x,y)  or (-x,y) or (-x,-y) or ( x,-y) reason being that you haven't given point O lies in which quadrant.

Rotation through 90° counter clockwise

(x,y) = (-y,x)

(-x,y)=(-y,-x)

(-x,-y)=(y,-x)

(x,-y) =(y,x)

Then Reflection across X axis has taken place.

(-y,x) = (-y,-x)

(-y,-x)=(-y,x)

(y,-x)=(y,x)

(y,x)=(y,-x)

Then a reflection across the y-axis has taken place.

(-y,-x)=(y,-x)

(-y,x) = (y,x)

(y,x) =(-y,x)

(y,-x)=(-y,-x)

Then a translation a units to the right and b units up has taken place.

(y,-x)=(y+a,-x+b)

(y,x) =(y+a, x+b)

(-y,x) =(-y+a,x+b)

(-y,-x)=(-y+a,-x+b)

Then a rotation of 180 degrees counterclockwise about the origin has taken place.

(y+a,-x+b)=[-(y+a),-(-x+b)]

(y+a, x+b)=[- (y+a), -(x+b)]

(-y+a,x+b)=[-(-y+a),-(x+b)]

(-y+a,-x+b) =[-(-y+a),-(-x+b)]

Now Again a reflection across the Y axis has taken place.

[-(y+a),-(-x+b)]=[(y+a),-(-x+b)]

[-(y+a),-(x+b)]=[(y+a),-(-x+b)]

[-(-y+a),-(x+b)]=[(-y+a),-(x+b)]

[-(-y+a),-(-x+b)]=[(-y+a),-(-x+b)]

Totally depends on value of a and b on which quadrant these point lies.



7 0
3 years ago
Claire Judice
Julli [10]

Answer:

The values of 'x' are -1.2, 0, 0, -4i or 4i.

Step-by-step explanation:

Given:

The equation to solve is given as:

5x^5+6x^4+80x^3+96x^2=0

Factoring x^2 from all the terms, we get:

x^2(5x^3+6x^2+80x+96)=0

Now, rearranging the terms, we get:

x^2(5x^3+80x+6x^2+96)=0

Now, factoring 5x from the first two terms and 6 from the last two terms, we get:

x^2(5x(x^2+16)+6(x^2+16))=0\\x^2(x^2+16)(5x+6)=0

Now, equating each factor to 0 and solving for 'x', we get:

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There are 3 real values and 2 imaginary values. The value of 'x' as 0 is repeated twice.

Therefore, the values of 'x' are -1.2, 0, 0, -4i or 4i.

4 0
4 years ago
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